Storing vectors of different lengths

Dear Everyone,

Just a quick question:

I have 4 vectors, each of different lengths, and I want to store them for some calculations later on in my script.

So I tried:

Hx = [36 72 108 144]  % Height of building
v = Hx./3*15 % Maximum shear force acting on the building

V = flip((15:15:v))   % 4 vectors containing the cumulative shear forces at each storey

But colon operators must be real scalars. So, this won't work.

My question is: what is the simplest way to obtain and store the 4 vectors I want of varying lengths?

Many thanks to you all.

Scott

[Scott Banks]

Thank you, guys, that's great!

Answer 1

[dpb]

Hx = [36 72 108 144]  % Height of building

Hx = 1×4

36 72 108 144

<mw-icon class=""></mw-icon>

<mw-icon class=""></mw-icon>

v = Hx./3*15 % Maximum shear force acting on the building

v = 1×4

180 360 540 720

<mw-icon class=""></mw-icon>

<mw-icon class=""></mw-icon>

V = arrayfun(@(h)[15:15:h].',v,'uni',0);

maybe is what you had in mind?

The above will be the same for each up to the preceding length since set a fixed lower limit and step size for each; maybe you intended to use a zero for the first and then the height of the previous story for the base of the next?

That would be something like

V = arrayfun(@(f,c)[f:15:c].',[0 v(1:end-1)],v,'uni',0)

V = 1×4 cell array
    {13×1 double}    {13×1 double}    {13×1 double}    {13×1 double}

[Scott Banks]

Hi dpb, thanks for that. however, I need the vectors flipped so that the shears are descending. For example, the first vector goes [15, 30, 45 ..... 180]. I need [180, 165 150.... 15].

[Torsten]

V = arrayfun(@(h)flipud([15:15:h].'),v,'uni',0)

or

V = arrayfun(@(f,c)flipud([f:15:c].'),[0 v(1:end-1)],v,'uni',0)

[Stephen23]

Without the misleading square brackets:

Hx = [36,72,108,144];
v = Hx./3*15;
V = arrayfun(@(h)flip(15:15:h),v, 'uni',0);

Checking:

V{:}

ans = 1×12

180 165 150 135 120 105 90 75 60 45 30 15

<mw-icon class=""></mw-icon>

<mw-icon class=""></mw-icon>

ans = 1×24

360 345 330 315 300 285 270 255 240 225 210 195 180 165 150 135 120 105 90 75 60 45 30 15

<mw-icon class=""></mw-icon>

<mw-icon class=""></mw-icon>

ans = 1×36

540 525 510 495 480 465 450 435 420 405 390 375 360 345 330 315 300 285 270 255 240 225 210 195 180 165 150 135 120 105

<mw-icon class=""></mw-icon>

<mw-icon class=""></mw-icon>

ans = 1×48

720 705 690 675 660 645 630 615 600 585 570 555 540 525 510 495 480 465 450 435 420 405 390 375 360 345 330 315 300 285

<mw-icon class=""></mw-icon>

<mw-icon class=""></mw-icon>

Answer 2

[Umar]

Hi @Scott Banks,

I saw your question about generating four vectors of varying lengths for building shear forces. I put together a clean MATLAB approach that does exactly what you need.

%% Building heights and max shear
Hx = [36, 72, 108, 144];
v  = Hx/3*15;
step = 15;

%% Generate all descending vectors in a single, compact line
V = arrayfun(@(x) linspace(x, step, ceil(x/step)), v, 'UniformOutput', false);

%% Display each vector
cellfun(@(vec,i) fprintf('Building %d | H = %d m | Shear: %s\n', i, Hx(i),   strjoin(string(vec), ' ')), V, num2cell(1:numel(V)));

Results: please see attached.

How it works:

• Each element of `V` is a descending vector from the maximum shear down to 15 kN.

• The vectors automatically have the correct length for each building.

• All vectors are stored in a cell array, making them easy to access later (`V{1}`, `V{2}`, etc.).

• No loops are needed — the code is concise, readable, and scalable.

This approach directly gives you the four vectors you wanted, in a neat, structured way. It’s also easy to extend if you add more buildings or change the step size.

Hope this helps!